## Description

https://leetcode.com/problems/word-search/

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

**Example:**

board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] Given word = "ABCCED", returntrue. Given word = "SEE", returntrue. Given word = "ABCB", returnfalse.

**Constraints:**

`board`

and`word`

consists only of lowercase and uppercase English letters.`1 <= board.length <= 200`

`1 <= board[i].length <= 200`

`1 <= word.length <= 10^3`

## Explanation

The solution would be backtracking, which is a methodology where we mark the current path of exploration, if the path does not lead to a solution, we then revert the change (i.e. backtracking) and try another path.

As the general idea for the solution, we would walk around the 2D grid, at each step we *mark* our choice before jumping into the next step. And at the end of each step, we would also revert our marking, so that we could have a *clean slate* to try another *direction*. In addition, the exploration is done via the *DFS* strategy, where we go as further as possible before we try the next direction.

## Python Solution

```
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
if (len(board) == 0 or len(board[0]) == 0):
return False
for i in range(0, len(board)):
for j in range(0, len(board[0])):
if self.helper(board, i, j, word):
return True
return False
def helper(self, board, i, j, word):
if len(word) == 0:
return True
if i < 0 or i > len(board) - 1 or j < 0 or j > len(board[0]) - 1:
return False
if board[i][j] != word[0]:
return False
char = board[i][j]
board[i][j] = '#'
suffix = word[1:]
result = self.helper(board, i + 1, j, suffix) or self.helper(board, i - 1, j, suffix) or self.helper(board, i, j + 1, suffix) or self.helper(board, i, j - 1, suffix)
board[i][j] = char
return result
```

- Time complexity: O(N * 4^L). where N is the number of cells in the board and L is the length of the word to be matched. For the backtracking function, its execution trace would be visualized as a 4-ary tree, each of the branches represent a potential exploration in the corresponding direction. Therefore, in the worst case, the total number of invocation would be the number of nodes in a full 4-nary tree, which is about 4^L. We iterate through the board for backtracking,
*i.e.*there could be N times invocation for the backtracking function in the worst case. As a result, overall the time complexity of the algorithm would be O(N * 4^L).

- Space complexity: O(L). where L is the length of the word to be matched.

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